pH, pOH, and the pH scale (article)

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Definitions of pH, pOH, and the pH scale. Calculating the pH of a strong acid or base solution. The relationship between acid strength and the pH of a solution.

Key points

We can convert between $[H^{+}]$ ‍ and $pH$ ‍ using the following equations:

$\begin{aligned} pH & = - \log [H^{+}] \\ [H^{+}] & = 10^{- pH} \end{aligned}$ ‍

We can convert between $[{OH}^{-}]$ ‍ and $pOH$ ‍ using the following equations:

$\begin{aligned} pOH & = - \log [{OH}^{-}] \\ [{OH}^{-}] & = 10^{- pOH} \end{aligned}$ ‍

For any aqueous solution at $25^{\circ} C$ ‍:

$pH + pOH = 14$ ‍.

For every factor of $10$ ‍ increase in concentration of $[H^{+}]$ ‍, $pH$ ‍ will decrease by $1$ ‍ unit, and vice versa.
Both acid strength and concentration determine $[H^{+}]$ ‍ and $pH$ ‍.

Introduction

In aqueous solution, an acid is defined as any species that increases the concentration of $H^{+} (a q)$ ‍, while a base increases the concentration of ${OH}^{-} (a q)$ ‍. Typical concentrations of these ions in solution can be very small, and they also span a wide range.

For example, a sample of pure water at $25^{\circ} C$ ‍ contains $1.0 \times 10^{- 7} M$ ‍ of $H^{+}$ ‍ and ${OH}^{-}$ ‍. In comparison, the concentration of $H^{+}$ ‍ in stomach acid can be up to approximately $1.0 \times 10^{- 1} M$ ‍. That means $[H^{+}]$ ‍ for stomach acid is approximately $6$ ‍ orders of magnitude larger than in pure water!

To avoid dealing with such hairy numbers, scientists convert these concentrations to $pH$ ‍ or $pOH$ ‍ values. Let's look at the definitions of $pH$ ‍ and $pOH$ ‍.

Definitions of $pH$ ‍ and $pOH$ ‍

Relating $[H^{+}]$ ‍ and $pH$ ‍

The $pH$ ‍ for an aqueous solution is calculated from $[H^{+}]$ ‍ using the following equation:

$pH = - \log [H^{+}] (Eq. 1a)$ ‍

The lowercase $p$ ‍ indicates $‘ ‘ - {log}_{10} "$ ‍. You will often see people leave off the base $10$ ‍ part as an abbreviation.

For example, if we have a solution with $[H^{+}] = 1 \times 10^{- 5} M$ ‍, then we can calculate the $pH$ ‍ using Eq. 1a:

$pH = - \log (1 \times 10^{- 5}) = 5.0$ ‍

Given the $pH$ ‍ of a solution, we can also find $[H^{+}]$ ‍:

$[H^{+}] = 10^{- pH} (Eq. 1b)$ ‍

There are a few rules you want to know for calculations involving logarithms like $pH$ ‍ and $pOH$ ‍. For a review of log properties, see Sal's video on properties of logarithms.

Relating $[{OH}^{-}]$ ‍ and $pOH$ ‍

The $pOH$ ‍ for an aqueous solution is defined in the same way for $[{OH}^{-}]$ ‍:

$pOH = - \log [{OH}^{-}] (Eq. 2a)$ ‍

For example, if we have a solution with $[{OH}^{-}] = 1 \times 10^{- 12} M$ ‍, then we can calculate $pOH$ ‍ using Eq. 2a:

$pOH = - \log (1 \times 10^{- 12}) = 12.0$ ‍

Given the $pOH$ ‍ of a solution, we can also find $[{OH}^{-}]$ ‍:

$10^{- pOH} = [{OH}^{-}] (Eq. 2b)$ ‍

Relating $pH$ ‍ and $pOH$ ‍

Based on equilibrium concentrations of $H^{+}$ ‍ and ${OH}^{-}$ ‍ in water, the following relationship is true for any aqueous solution at $25^{\circ} C$ ‍:

$pH + pOH = 14 (Eq. 3)$ ‍

This relationship can be used to convert between $pH$ ‍ and $pOH$ ‍. In combination with Eq. 1a/b and Eq. 2a/b, we can always relate $pOH$ ‍ and/or $pH$ ‍ to $[{OH}^{-}]$ ‍ and $[H^{+}]$ ‍. For a derivation of this equation, check out the article on the autoionization of water.

Example 1: Calculating the $pH$ ‍ of a strong base solution

If we use $1.0 mmol$ ‍ of $NaOH$ ‍ to make $1.0 L$ ‍ of an aqueous solution at $25^{\circ} C$ ‍, what is the $pH$ ‍ of this solution?

We can find the $pH$ ‍ of our $NaOH$ ‍ solution by using the relationship between $[{OH}^{-}]$ ‍, $pH$ ‍, and $pOH$ ‍. Let's go through the calculation step-by-step.

Step 1. Calculate the molar concentration of $NaOH$ ‍

Molar concentration is equal to moles of solute per liter of solution:

$Molar concentration = \frac{mol solute}{L solution}$ ‍

To calculate the molar concentration of $NaOH$ ‍, we can use the known values for the moles of $NaOH$ ‍ and the volume of solution:

$\begin{aligned} [NaOH] & = \frac{1.0 mmol NaOH}{1.0 L} \\ = \frac{1.0 \times 10^{- 3} mol NaOH}{1.0 L} \\ = 1.0 \times 10^{- 3} M NaOH \end{aligned}$ ‍

The concentration of $NaOH$ ‍ in the solution is $1.0 \times 10^{- 3} M$ ‍.

Step 2: Calculate $[{OH}^{-}]$ ‍ based on the dissociation of $NaOH$ ‍

Because $NaOH$ ‍ is a strong base, it dissociates completely into its constituent ions in aqueous solution:

$NaOH (a q) \to {Na}^{+} (a q) + {OH}^{-} (a q)$ ‍

This balanced equation tells us that every mole of $NaOH$ ‍ produces one mole of ${OH}^{-}$ ‍ in aqueous solution. Therefore, we have the following relationship between $[NaOH]$ ‍ and $[{OH}^{-}]$ ‍:

$[NaOH] = [{OH}^{-}] = 1.0 \times 10^{- 3} M$ ‍

Step 3: Calculate $pOH$ ‍ from $[{OH}^{-}]$ ‍ using Eq. 2a

Now that we know the concentration of ${OH}^{-}$ ‍, we can calculate $pOH$ ‍ using Eq. 2a:

$\begin{aligned} pOH & = - \log [{OH}^{-}] \\ = - \log (1.0 \times 10^{- 3}) \\ = 3.00 \end{aligned}$ ‍

The $pOH$ ‍ of our solution is $3.00$ ‍.

Step 4: Calculate $pH$ ‍ from $pOH$ ‍ using Eq. 3

We can calculate $pH$ ‍ from $pOH$ ‍ using Eq. 3. Rearranging to solve for our unknown, $pH$ ‍:

$pH = 14 - pOH$ ‍

We can substitute the value of $pOH$ ‍ we found in Step 3 to find the $pH$ ‍:

$pH = 14 - 3.00 = 11.00$ ‍

Therefore, the $pH$ ‍ of our $NaOH$ ‍ solution is $11.00$ ‍.

The $pH$ ‍ scale: Acidic, basic, and neutral solutions

Converting $[H^{+}]$ ‍ to $pH$ ‍ is a convenient way to gauge the relative acidity or basicity of a solution. The $pH$ ‍ scale allows us to easily rank different substances by their $pH$ ‍ value.

The $pH$ ‍ scale is a negative logarithmic scale. The logarithmic part means that $pH$ ‍ changes by $1$ ‍ unit for every factor of $10$ ‍ change in concentration of $H^{+}$ ‍. The negative sign in front of the log tells us that there is an inverse relationship between $pH$ ‍ and $[H^{+}]$ ‍: when $pH$ ‍ increases, $[H^{+}]$ ‍ decreases, and vice versa.

The following image shows a $pH$ ‍ scale labeled with $pH$ ‍ values for some common household substances. These $pH$ ‍ values are for solutions at $25^{\circ} C$ ‍. Note that it is possible to have a negative $pH$ ‍ value.

Some important terminology to remember for aqueous solutions at $25^{\circ} C$ ‍:

For a neutral solution, $pH = 7$ ‍.
Acidic solutions have $pH < 7$ ‍.
Basic solutions have $pH > 7$ ‍.

The lower the $pH$ ‍ value, the more acidic the solution and the higher the concentration of $H^{+}$ ‍. The higher the $pH$ ‍ value, the more basic the solution and the lower the concentration of $H^{+}$ ‍. While we could also describe the acidity or basicity of a solution in terms of $pOH$ ‍, it is a little more common to use $pH$ ‍. Luckily, we can easily convert between $pH$ ‍ and $pOH$ ‍ values.

Concept check: Based on the $pH$ ‍ scale given above, which solution is more acidic $-$ ‍orange juice, or vinegar?

Vinegar has a $pH$ ‍ of $\approx 3.0$ ‍, and orange juice has a $pH$ ‍ of $\approx 3.6$ ‍. Because the $pH$ ‍ of vinegar is less than that of orange juice $(3.0 < 3.6)$ ‍, vinegar has a higher concentration of $H^{+}$ ‍.

Therefore, vinegar is more acidic than orange juice.

Example $2$ ‍: Determining the $pH$ ‍ of a diluted strong acid solution

We have $100 mL$ ‍ of a nitric acid solution with a $pH$ ‍ of $4.0$ ‍. We dilute the solution by adding water to get a total volume of $1.0 L$ ‍.

What is the $pH$ ‍ of the diluted solution?

There are multiple ways to solve this problem. We will go over two different methods.

Method 1. Use properties of the log scale

Recall that $pH$ ‍ scale is a negative logarithmic scale. Therefore, if the concentration of $H^{+}$ ‍ decreases by a single factor of $10$ ‍, then the $pH$ ‍ will increase by $1$ ‍ unit.

Since the original volume, $100 mL$ ‍, is one tenth the total volume after dilution, the concentration of $H^{+}$ ‍ in solution has been reduced by a factor of $10$ ‍. Therefore, the $pH$ ‍ of the solution will increase $1$ ‍ unit:

$\begin{aligned} pH & = original pH + 1.0 \\ = 4.0 + 1.0 \\ = 5.0 \end{aligned}$ ‍
See Also
What is the pH scale and what does it measure? - BBC Bitesize pH Scale: Acids, bases, pH and buffers (article) | Khan Academy

Therefore, the $pH$ ‍ of the diluted solution is $5.0$ ‍.

Method 2. Use moles of $H^{+}$ ‍ to calculate $pH$ ‍

Step 1: Calculate moles of $H^{+}$ ‍

We can use the $pH$ ‍ and volume of the original solution to calculate the moles of $H^{+}$ ‍ in the solution.

$\begin{aligned} {moles H}^{+} & = [H^{+}]_{i n i t i a l} \times volume \\ = 10^{- pH} M \times volume \\ = 10^{- 4.0} M \times 0.100 L \\ = 1.0 \times 10^{- 5} {mol H}^{+} \end{aligned}$ ‍

Step 2: Calculate molarity of $H^{+}$ ‍ after dilution

The molarity of the diluted solution can be calculated by using the moles of $H^{+}$ ‍ from the original solution and the total volume after dilution.

$\begin{aligned} [H^{+}]_{f i n a l} & = \frac{{mol H}^{+}}{L solution} \\ = \frac{1.0 \times 10^{- 5} {mol H}^{+}}{1.0 L} \\ = 1.0 \times 10^{- 5} M \end{aligned}$ ‍

Step 3: Calculate $pH$ ‍ from $[H^{+}]$ ‍

Finally, we can use Eq. 1a to calculate $pH$ ‍:

$\begin{aligned} pH & = - \log [H^{+}] \\ = - log (1.0 \times 10^{- 5}) \\ = 5.0 \end{aligned}$ ‍

Method 2 gives us the same answer as Method 1, hooray!

In general, Method 2 takes a few extra steps, but it can always be used to find changes in $pH$ ‍. Method 1 is a handy shortcut when changes in concentration occur as multiples of $10$ ‍. Method 1 can also be used as a quick way to estimate $pH$ ‍ changes.

Relationship between $pH$ ‍ and acid strength

Based on the equation for $pH$ ‍, we know that $pH$ ‍ is related to $[H^{+}]$ ‍. However, it is important to remember that $pH$ ‍ is not always directly related to acid strength.

The strength of an acid depends on the amount that the acid dissociates in solution: the stronger the acid, the higher $[H^{+}]$ ‍ at a given acid concentration. For example, a $1.0 M$ ‍ solution of strong acid $HCl$ ‍ will have a higher concentration of $H^{+}$ ‍ than a $1.0 M$ ‍ solution of weak acid $HF$ ‍. Thus, for two solutions of monoprotic acid at the same concentration, $pH$ ‍ will be proportional to acid strength.

More generally though, both acid strength and concentration determine $[H^{+}]$ ‍. Therefore, we can't always assume that the $pH$ ‍ of a strong acid solution will be lower than the $pH$ ‍ of a weak acid solution. The acid concentration matters too!

Summary

We can convert between $[H^{+}]$ ‍ and $pH$ ‍ using the following equations:

$\begin{aligned} pH & = - \log [H^{+}] \\ [H^{+}] & = 10^{- pH} \end{aligned}$ ‍

We can convert between $[{OH}^{-}]$ ‍ and $pOH$ ‍ using the following equations:

$\begin{aligned} pOH & = - \log [{OH}^{-}] \\ [{OH}^{-}] & = 10^{- pOH} \end{aligned}$ ‍

For every factor of $10$ ‍ increase in concentration of $[H^{+}]$ ‍, $pH$ ‍ will decrease by $1$ ‍ unit, and vice versa.
For any aqueous solution at $25^{\circ} C$ ‍:

$pH + pOH = 14$ ‍.

Both acid strength and concentration determine $[H^{+}]$ ‍ and $pH$ ‍.

Attributions

This article was adapted from the following articles:

“The pH Scale” from UC Davis ChemWiki, CC BY-NC-SA 3.0

The modified article is licensed under a CC-BY-NC-SA 4.0 license.

Additional References

Zumdahl, S.S., and Zumdahl S.A. (2003). Atomic Structure and Periodicity. In Chemistry (6th ed., pp. 290-94), Boston, MA: Houghton Mifflin Company.

Problem 1: Calculating the pH of a strong base solution at $25^{\circ} C$ ‍

We make $200 mL$ ‍ of a solution with a $0.025 M$ ‍ concentration of ${Ca(OH)}_{2}$ ‍. The solution is then diluted to $1.00 L$ ‍ by adding additional water.

What is the $pH$ ‍ of the solution after dilution?

Choose 1 answer:

(Choice A)
$2.00$ ‍
(Choice B)
$13.00$ ‍
(Choice C)
$11.00$ ‍
(Choice D)
$12.00$ ‍
(Choice E)
$3.00$ ‍

Calcium hydroxide dissociates in aqueous solution as follows:

${Ca(OH)}_{2} (a q) \to {Ca}^{2 +} (a q) + 2 {OH}^{-} (a q)$ ‍

We see that $2 {mol OH}^{-}$ ‍ are produced for every $1 {mol Ca(OH)}_{2}$ ‍ that dissociates. Therefore, the concentration of hydroxide ion, $[{OH}^{-}]$ ‍, will be twice the molarity of ${Ca(OH)}_{2}$ ‍:

$[{OH}^{-}]_{initial} = 0.025 M \times \frac{2 {mol OH}^{-}}{1 {mol Ca(OH)}_{2}} = 0.050 M$ ‍

The solution has been diluted by a factor of $5$ ‍, since $200 mL \times 5 = 1.0 L$ ‍. Therefore, the final molar concentration of ${OH}^{-}$ ‍ will be $\frac{1}{5}$ ‍ its initial value:

$[{OH}^{-}]_{final} = 0.050 M \times \frac{1}{5} = 0.010 M$ ‍

We can calculate the $pOH$ ‍ of the diluted solution from $[{OH}^{-}]$ ‍:

$pOH = - \log [{OH}^{-}] = - \log (0.010) = 2.00$ ‍

Since $pH + pOH = 14.00$ ‍ at $25^{\circ} C$ ‍,

$\begin{aligned} pH & = 14.00 - pOH \\ = 14.00 - 2.00 \\ = 12.00 \end{aligned}$ ‍

Therefore, the $pH$ ‍ of the diluted solution is $12.00$ ‍.

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Jonathan Ziesmer
8 years agoPosted 8 years ago. Direct link to Jonathan Ziesmer's post “How does the temperature ...”
How does the temperature affect the pH and pOH?
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(37 votes)
- Matt B
  8 years agoPosted 8 years ago. Direct link to Matt B's post “At 100C the pH of water i...”
  At 100C the pH of water is 6.14, so higher temperature decreases the pH. The opposite is true for pOH: higher temperatures increases the pOH.
  Comment on Matt B's post “At 100C the pH of water i...”
  (44 votes)
Kaylee Wilson
8 years agoPosted 8 years ago. Direct link to Kaylee Wilson's post “What does M stand for in ...”
What does M stand for in the unit labels?
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(15 votes)
- SRIVATHSAN B
  8 years agoPosted 8 years ago. Direct link to SRIVATHSAN B's post “M stands for the unit of ...”
  M stands for the unit of Molarity of the solution.
  Comment on SRIVATHSAN B's post “M stands for the unit of ...”
  (20 votes)
Amit Mukherjee
8 years agoPosted 8 years ago. Direct link to Amit Mukherjee's post “Could someone explain the...”
Could someone explain the difference between acid strength and concentration? According to me, a strong acid will fully ionise in water compared to a weak acid which will partially ionise. Therefore a strong acid will contribute more H+ ions than a weak acid. Therefore, the pH of a strong acid solution will be higher than a weak acid solution.
Is this correct?
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(8 votes)
- Matt B
  8 years agoPosted 8 years ago. Direct link to Matt B's post “Nice question!! It is imp...”
  Nice question!! It is important that you don't confuse the words strong and weak with the terms concentrated and dilute. At the same concentration, a weak acid will be less acidic than a strong acid. However, if you have highly concentrated weak acid (almost pure) and compare this to a very diluted strong acid (like 1 drop of HCl in a swimming pool) then the pH of the weak acid will be much more acidic than that of the strong acid.
  Comment on Matt B's post “Nice question!! It is imp...”
  (25 votes)
SRIVATHSAN B
8 years agoPosted 8 years ago. Direct link to SRIVATHSAN B's post “Can someone please explai...”
Can someone please explain what are monoprotic and diprotic acids? Thanks.
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(13 votes)
- Davide Ghazal
  8 years agoPosted 8 years ago. Direct link to Davide Ghazal's post “H2SO4 is a typical diprot...”
  H2SO4 is a typical diprotic acid (2 protons can be released in aqueous solution, however one at that time)
  H2SO4 +H20 gives HSO4- + H+/H3O+
  then,
  HSO4- +H2O gives SO42- + H+/H3O+
  Comment on Davide Ghazal's post “H2SO4 is a typical diprot...”
  (16 votes)
areebaali160
8 years agoPosted 8 years ago. Direct link to areebaali160's post “how can we solve pH,pOH n...”
how can we solve pH,pOH numericals without using scientific calculator during our examination?
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(8 votes)
- Matt B
  8 years agoPosted 8 years ago. Direct link to Matt B's post “Unless you are using perf...”
  Unless you are using perfect numbers of base ten (e.g. 10^-7, 10^-2, etc.) there is no way to do it because you cannot easily do logarithms. If you are asked to do these calculations without a calculator, there is a good chance minimal if any extensive calculations are required.
  Example: log(10^-6) = -6
  Comment on Matt B's post “Unless you are using perf...”
  (7 votes)
apuri
8 years agoPosted 8 years ago. Direct link to apuri's post “How can NaOH have a pH sc...”
How can NaOH have a pH scale? How can a base add H+ ions to the solution? It adds OH- ions right?
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(5 votes)
- Ernest Zinck
  8 years agoPosted 8 years ago. Direct link to Ernest Zinck's post “NaOH does not have a pH, ...”
  NaOH does not have a pH, but an aqueous solution of NaOH does.
  Water contains both H⁺ and OH⁻ ions.
  Adding NaOH increases the concentration of OH⁻ ions and decreases the concentration of H⁺ ions.
  But there are always some H⁺ ions present, so aqueous NaOH solutions have a pH, usually between 7 and 14.
  Comment on Ernest Zinck's post “NaOH does not have a pH, ...”
  (11 votes)
Michelle
8 years agoPosted 8 years ago. Direct link to Michelle's post “How does pH+pOH= 14? Wher...”
How does pH+pOH= 14? Where does the random number come in?
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(3 votes)
- Ernest Zinck
  8 years agoPosted 8 years ago. Direct link to Ernest Zinck's post “There is no random number...”
  There is no random number.
  The formula comes from the ion product for water.
  [H⁺][OH⁻] = 1.00 × 10⁻¹⁴
  ∴ pH + pOH = 14.00
  Comment on Ernest Zinck's post “There is no random number...”
  (4 votes)
m pe
6 years agoPosted 6 years ago. Direct link to m pe's post “HiI feel like there's a ...”
Hi
I feel like there's a step missing.I'm not sure why the pH as an exponent is negative & where the minus sign comes from.I understand that the logarithm (of base 10) was changed to an exponent.What is this law of logarithms called?
pH= -log (H+)
10^-pH = (H+)
Also I was trying to figure it out with numbers pH=-(log 10^-4)
and I got 10^-pH =10^-4 and I'm not sure where to go from there to obtain the pH.Do I just cancel out the 10s & minuses that are on both sides to get a pH of 4,to cross off these I have to divide/multiply both sides by some number(s) would these numbers be 10 and multiplying the exponents by -1 to get rid of the minuses because the pH scale is usually positive numbers?
Thanks!
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(3 votes)
- RogerP
  6 years agoPosted 6 years ago. Direct link to RogerP's post “This is the _power rule_ ...”
  This is the power rule of logs. When you have a number in front of a log term, this is the same as raising the log term to that number. For example, 4log(3) is the same as log(3^4).
  With pH, the number in front of the log is -1 (because pH = -log [H+]). Therefore, using the power rule, we can re-express this as pH = log ([H+]^-1).
  Using another log rule, we can express each side of this equation as an exponent of 10 and we get:
  10^pH = 10^log ([H+]^-1).
  Using a definition of logs, the right hand side of this equation now just becomes [H+]^-1. So we have:
  10^pH = [H+]^-1
  The right hand side can be expressed as 1/[H+] giving us:
  10^pH = 1/[H+]
  Multiplying each side by [H+] and dividing each side by 10^pH gives:
  [H+] = 1/10^pH which is the same as saying [H+] = 10^-pH.
  As an example, if the pH is 7, then [H+] = 10^-7.
  Comment on RogerP's post “This is the _power rule_ ...”
  (3 votes)
Nicolas Jaramillo
7 years agoPosted 7 years ago. Direct link to Nicolas Jaramillo's post “what is -log? is it a num...”
what is -log? is it a number?
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(2 votes)
- Ryan W
  7 years agoPosted 7 years ago. Direct link to Ryan W's post “Log is a maths function, ...”
  Log is a maths function, it stands for logarithm.
  We use a log scale so we can visualise the concentration of H+ ions easier as we don't without have to deal with lots of decimal places etc.
  Button navigates to signup page
  (4 votes)
Aditya Shelke
6 years agoPosted 6 years ago. Direct link to Aditya Shelke's post “Is it necessary that to c...”
Is it necessary that to calculate pH or pOH of a substance, we need to have the substance in aqueous physical state? What if the substance is already in aqueous state?
Is it possible to find the pH or pOH of a substance in solid physical state or even by melting it?
Also, we always calculate the pH or pOH of a solid substance by dissolving it in water and find its pH or pOH according to the [H+] or [OH-] produced by it .Why we don't take in consideration the [H+] or [OH-] present in water and only take [H+] or [OH-]of the dissolved substance?
Thankyou :)
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(3 votes)
- Shlok Sanju
  7 months agoPosted 7 months ago. Direct link to Shlok Sanju's post “Ok so while calculating t...”
  Ok so while calculating the equilibrium constant for a reaction, we consider the active masses of substances involved in the reactants and products. Since the activity of pure solids and liquids is unity, adding them to the expression will make no difference and it is usually omitted.
  Hope this helps!
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  (1 vote)

pH, pOH, and the pH scale (article) | Khan Academy (2024)

Key points

Introduction

Definitions of pH‍ and pOH‍

Relating [H+]‍ and pH‍

Relating [OH−]‍ and pOH‍

Relating pH‍ and pOH‍

Example 1: Calculating the pH‍ of a strong base solution

Step 1. Calculate the molar concentration of NaOH‍

Step 2: Calculate [OH−]‍ based on the dissociation of NaOH‍

Step 3: Calculate pOH‍ from [OH−]‍ using Eq. 2a

Step 4: Calculate pH‍ from pOH‍ using Eq. 3

The pH‍ scale: Acidic, basic, and neutral solutions

Example 2‍: Determining the pH‍ of a diluted strong acid solution